Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

2. A criminologist developed a test to measure recidivism, where low scores indi

ID: 3060234 • Letter: 2

Question

2. A criminologist developed a test to measure recidivism, where low scores indicated a lower probability of repeating the undesirable behavior. The test is normal so it has a mean of 140 and a standard deviation of 40.
(a) What is the percentile rank of a score of 172?
(b) What is the Z score for a test score of 200?
(c) What percentage of scores falls between 100 and 160?
(d) What proportion of respondents should score above 190?
(e) Suppose an individual is in the 67th percentile in this test, what is his/her corresponding recidivism score? 2. A criminologist developed a test to measure recidivism, where low scores indicated a lower probability of repeating the undesirable behavior. The test is normal so it has a mean of 140 and a standard deviation of 40.
(a) What is the percentile rank of a score of 172?
(b) What is the Z score for a test score of 200?
(c) What percentage of scores falls between 100 and 160?
(d) What proportion of respondents should score above 190?
(e) Suppose an individual is in the 67th percentile in this test, what is his/her corresponding recidivism score? 2. A criminologist developed a test to measure recidivism, where low scores indicated a lower probability of repeating the undesirable behavior. The test is normal so it has a mean of 140 and a standard deviation of 40.
(a) What is the percentile rank of a score of 172?
(b) What is the Z score for a test score of 200?
(c) What percentage of scores falls between 100 and 160?
(d) What proportion of respondents should score above 190?
(e) Suppose an individual is in the 67th percentile in this test, what is his/her corresponding recidivism score?

Explanation / Answer

a)

z score =(172-140)/10 =0.8

for above percentile rank =78.81  

b)

z score =(200-140)/40 =1.5

c)

d)

e)

for 67th percentile z score =0.44

therefore corresponding score =mean+z*std deviation =140+0.44*40 =157.60

for normal distribution z score =(X-)/ here mean=       = 140 std deviation   == 40.0000